Question: $\dfrac{ -3d + e }{ -3 } = \dfrac{ -7d + 2f }{ -10 }$ Solve for $d$.
Answer: Multiply both sides by the left denominator. $\dfrac{ -3d + e }{ -{3} } = \dfrac{ -7d + 2f }{ -10 }$ $-{3} \cdot \dfrac{ -3d + e }{ -{3} } = -{3} \cdot \dfrac{ -7d + 2f }{ -10 }$ $-3d + e = -{3} \cdot \dfrac { -7d + 2f }{ -10 }$ Multiply both sides by the right denominator. $-3d + e = -3 \cdot \dfrac{ -7d + 2f }{ -{10} }$ $-{10} \cdot \left( -3d + e \right) = -{10} \cdot -3 \cdot \dfrac{ -7d + 2f }{ -{10} }$ $-{10} \cdot \left( -3d + e \right) = -3 \cdot \left( -7d + 2f \right)$ Distribute both sides $-{10} \cdot \left( -3d + e \right) = -{3} \cdot \left( -7d + 2f \right)$ ${30}d - {10}e = {21}d - {6}f$ Combine $d$ terms on the left. ${30d} - 10e = {21d} - 6f$ ${9d} - 10e = -6f$ Move the $e$ term to the right. $9d - {10e} = -6f$ $9d = -6f + {10e}$ Isolate $d$ by dividing both sides by its coefficient. ${9}d = -6f + 10e$ $d = \dfrac{ -6f + 10e }{ {9} }$